ICSE Class 10 Maths Geometry Loci – Concepts & Theorems

What is a Locus?

Imagine you have a pet dog that loves to stay exactly 5 metres away from a big oak tree. If you draw every possible spot where the dog could stand, you’d get a perfect circle around the tree. That collection of points is called a locus – the set of all points that satisfy a given condition.

💡 In Simple Words: A locus is just a fancy word for “all the places something can be” when it follows a rule. Think of it as the trail of dots you’d get if you walked around while keeping the same distance from a fixed spot.

How to Find a Locus – Step by Step

When a question asks you to “find the locus of…”, follow this short recipe:

Each step turns a word problem into an algebraic equation you can recognise.

Common Locus Theorems You Must Know

Worked Example 1 – Circle Locus

Find the locus of a point P(x, y) that stays 4 cm away from the origin O(0,0).

1. Write the distance condition: OP = 4.
2. Replace OP with the distance formula √(x²+y²) = 4.
3. Square both sides to remove the root: x² + y² = 16.
4. This is the standard form of a circle centred at the origin with radius 4 cm.

So the answer is a circle with equation x² + y² = 16.

Worked Example 2 – Perpendicular Bisector

Find the locus of points that are equally distant from A(2,‑1) and B(‑2,3).

1. Let P be (x, y). Write PA = PB.
2. Use the distance formula: √[(x‑2)²+(y+1)²] = √[(x+2)²+(y‑3)²].
3. Square both sides and simplify:
   • (x‑2)²+(y+1)² = (x+2)²+(y‑3)²
   • Expand: x²‑4x+4 + y²+2y+1 = x²+4x+4 + y²‑6y+9
   • Cancel x² and y², bring terms together: -4x+2y+5 = 4x‑6y+13
   • Combine: -8x+8y = 8 → y = x + 1.
4. Result: a straight line y = x + 1, which is the perpendicular bisector of AB.

Tips for Solving Locus Problems Quickly

• Translate words into equations early. “Fixed distance” → use the distance formula.
• Watch out for “twice the distance” or “half the distance” – just multiply or divide r before squaring.
• If the condition involves a line, replace the perpendicular distance with the formula |Ax+By+C|/√(A²+B²).
• After simplifying, try to match the result with one of the common loci in the table above.
• Sketch a quick graph to check if your equation makes sense (e.g., radius should be positive).

📝 Likely Exam Questions

1. Question: Find the locus of a point P such that the sum of its distances from (‑3,0) and (3,0) is 10.
   Answer: The condition describes an ellipse with foci at (‑3,0) and (3,0) and major axis length 10. Hence the equation is \(\frac{x^2}{25}+\frac{y^2}{16}=1\).
2. Question: Determine the locus of points that are twice as far from the point (2,‑2) as from the line y = x.
   Answer: Using distance formulas, 2·√[(x‑2)²+(y+2)²] = |x‑y|/√2. Squaring and simplifying yields the parabola \((x‑2)^2+(y+2)^2 = \frac{(x‑y)^2}{8}\).
3. Question: Find the locus of a point whose distance from the origin is always three times its distance from the line y = 4.
   Answer: Let P(x,y). Condition: √(x²+y²) = 3·|y‑4|. Squaring gives x² + y² = 9(y‑4)², which simplifies to a circle after expansion: x² + y² = 9y²‑72y+144 → 8y²‑72y+144‑x² = 0, or x² = 8(y²‑9y+18). This is a circle after completing the square.
4. Question: The locus of a point P is the set of points equidistant from the points (1,2) and (5,6). Write its equation and describe its graph.
   Answer: Using the perpendicular bisector method, the equation reduces to y = x + 1, a straight line passing through (‑2,‑1) with slope 1.
5. Question: Prove that the locus of points whose distances from the axes are equal is the line y = x.
   Answer: Let the distances be |x| and |y|. Equality gives |x| = |y| → x² = y² → y = ±x. Since distances are non‑negative, the relevant locus in the first quadrant is y = x.